數例..邊個息做能幫幫嗎?~

Question

考慮200與500之間(孢括200和500),且滿呎下列條件的整數.然後求出它門之和 o

(a)可被7整除

(b)可被5及7整除

(打錯字)且滿足下列條件....

Answer 1

a )let T(n) = 7n

for 7n＞＝200 the smallest n = 29 T(29)=203
for 7n＜＝500 the largest n =71 T(71)=497
因此between 200 to 500，能被7整除的項數共 71-29+1=43
項差＝７，首項＝２０３，尾項＝４９７
所以總和＝（首項＋尾項）ｘ項差／２＝(203+497)x43/2=15050

b)總和(可被5及7整除)=總和(可被5整除)+總和(可被7整除)-總和(可被35整除)

for 5n＞＝200 the smallest n=40 T(40)=200
for 5n＜＝500 the largest n=100 T(100)=500
then 項數＝100 -40+1 =61
總和（可被５整除）＝(200+500)x61/2=21350

for 35n＞＝200, the smallest n =6 T(6)=210
for 35n＜＝500, the largest n=14 T(14)=490
then 項數=14-6+1=9
總和(可被35整除)=(210+490)x9/2=3150

therefore 總和(可被5及7整除)=15050+21350-3150=33250

2006-11-28 16:44:16 補充：
我將你(....及....)的意思看作or，若是and，答案就是總和(可被35整除)=3150

Answer 2

公式 : 1 + 2 + 3 + 4 + ...+ N = N ( N + 1) / 2

1) Consider : 200/7 = 28.571 500/7 = 71.428

Sum of all integer between 200 and 500 (可被 7 整除) =

7 ( 29 + 30 + ...+ 71) = 7 ((1 + 2 +..+ 71) - (1 + 2 + .. 28))
= 7 (71(71+1)/2 -28(28+1)/2)
= 7(2556 - 406)
= 15050
2) 可被 5 和 7 整除 --> 可被 35 整除

CONSIDER 200/35 = 5.71 500/35 = 14.2

Sum of all integer between 200 and 500 (可被 5 和 7 整除) =

35 ( 6 + 7 +... + 14) = 35 (( 1 + 2 + .. 14) - (1 + 2 + ..+5))
= 35 (14(14+1)/2 - 5(5+1)/2)
= 35 (105 - 15)
= 3150

Answer 3

a ) 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497.

b ) 210, 245, 280, 315, 350, 385, 420, 455, 490.